Question

A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent composition of fluorine

Answer 1

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

$percent composition element A=\frac{total mass of element A}{mass of compound} *100$

In this case, the percent composition of fluorine is:

$percent composition of fluorine=\frac{39.6 g}{60.3 g} *100$

percent composition of fluorine= 65.67%

The percent composition of fluorine is 65.67%